An excercise in graviation by misner, thorne and wheeler

[zasdzcc]

Homework Statement

My question is on problem 4.1 of Gravitation. In a generic case of electric field and magnetic field(i.e not $$E=0$$ or $$B=0$$ or $$E$$ and $$B$$ perpendicular), define the direction $$\hat{n}$$ unit vector ,

$$\hat{n}\tanh (2\alpha)=\frac{2\vec{E}\times\vec{B}}{\vec{E}^{2}+\vec{B}^2}$$

and $$\vec{\beta}=\tanh(\alpha)\hat{n}$$

is the velocity vector.

Show in the frame of the rocket with velocity $$\vec{\beta}$$, the Poynting vector vanishes.

Homework Equations

\begin{align}
\bar{\vec{E_{\parallel}}}&=\vec{E_{\parallel}}, \\
\bar{\vec{E_{\perp}}}&=\frac{\vec{E_{\perp}}+\vec{\beta}\times\vec{B_{\perp}}}{\sqrt{1-\beta^{2}}},\\
\bar{\vec{B_{\parallel}}}&=\vec{B_{\parallel}}, \\
\bar{\vec{B_{\perp}}}&=\frac{\vec{B_{\perp}}-\vec{\beta}\times\vec{E_{\perp}}}{\sqrt{1-\beta^{2}}}.
\end{align}
Note $$\beta\times X_{\perp}=\beta\times X$$

The Attempt at a Solution

I tried the following but I am stuck at the cancellation.

Let the $$\bar{\vec{E}}$$ and $$\bar{\vec{B}}$$ be the field in the rocket frame and the field without bars be the field of the rest frame. direction parallel along the velocity of rocket is denoted as subscript $$\parallel$$ and direction perpendicular to velocity of rocket direction is denoted by $$\perp$$ as the subscript.

By lorentz transformation.
\begin{align}
\bar{\vec{E_{\parallel}}}&=\vec{E_{\parallel}}, \\
\bar{\vec{E_{\perp}}}&=\frac{\vec{E_{\perp}}+\vec{\beta}\times\vec{B_{\perp}}}{\sqrt{1-\beta^{2}}},\\
\bar{\vec{B_{\parallel}}}&=\vec{B_{\parallel}}, \\
\bar{\vec{B_{\perp}}}&=\frac{\vec{B_{\perp}}-\vec{\beta}\times\vec{E_{\perp}}}{\sqrt{1-\beta^{2}}}.
\end{align}
Note $\beta\times X_{\perp}$=$\beta\times X$

In barred frame,
\begin{align}
\bar{\vec{E}}\times\bar{\vec{B}}
&=(\bar{\vec{E_{\perp}}}+\bar{\vec{E_{\parallel}}})\times(\bar{\vec{B_{\perp}}}+\bar{\vec{B_{\parallel}}})\\
&=\bar{\vec{E_{\perp}}}\times\bar{\vec{B_{\parallel}}}+\bar{\vec{E_{\parallel}}}\times\bar{\vec{B_{\perp}}}.
\end{align}
Plug in the lorentz transformation. One gets

$$\frac{\vec{E}\times\vec{B}-\vec{E}_{\parallel}\times(\vec{\beta}\times\vec{E})-\vec{B}_{\parallel}\times(\vec{\beta}\times\vec{B})}{\sqrt{1-\beta^{2}}}.$$

Now the last two terms looks generically like $$\vec{X_{\parallel}}\times(\vec{\beta}\times\vec{X})$$ simplifies to $$\vec{X_{\perp}}(\vec{X}\cdot\vec{\beta})$$.

However, I could not see the cancellation at this stage any where. Did I do something wrong?

 

[TSny]

Hello, zasdzcc.

How are ##\vec{E}## and ##\vec{B}## (in the original frame) oriented relative to the boost direction ##\vec{\beta}##? What does that tell you about ##\vec{E}_{\parallel}## and ##\vec{B}_{\parallel}##?

In barred frame,
\begin{align}
\bar{\vec{E}}\times\bar{\vec{B}}
&=(\bar{\vec{E_{\perp}}}+\bar{\vec{E_{\parallel}}})\times(\bar{\vec{B_{\perp}}}+\bar{\vec{B_{\parallel}}})\\
&=\bar{\vec{E_{\perp}}}\times\bar{\vec{B_{\parallel}}}+ \bar{\vec{E_{\parallel}}}\times\bar{\vec{B_{\perp}}}.
\end{align}

EDIT: You can't assume that ##\bar{\vec{E_{\perp}}} \times \bar{\vec{B_{\perp}}} = 0##. That's something you need to show.